Question

Suzanne deposits $3000 in an account that earns simple interest at an annual rate of 4.6%. Derek deposits $3000 in an account that eams compound interest at an annual rate of 4.6% and is compounded annually Complete the following table. (Round to the nearest dollar as needed.) Suzanne's Suzanne's Year Annual Interest and Balance Derek's Annual Interest and Derek's Balance.

Answer 1

Suzanne deposits $3000 in an account that earns simple interest at an annual rate of 4.6%. Derek deposits $3000 in an account that eams compound interest at an annual rate of 4.6% and is compounded annually

step 1

simple interest

we have that

The simple interest formula is equal to

`A=P(1+rt)`

`A=P\mleft(1+rt\mright)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

P=$3,000

r=4.6%=0.046

t=1 year

substitute in the formula

`\begin{gathered} A=3,000(1+0.046\cdot1) \\ A=\$3,138 \end{gathered}`

therefore

Suzanne's annual interest is (3,138-3,000)=$138

Suzanne's balance is $3,138

step 2

compound interest

we have that

The compound interest formula is equal to

`A=P(1+(r)/(n))^(nt)`

`A=P(1+(r)/(n))^(nt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

P=$3,000

r=4.6%=0.046

t=1 year

n=1

substitute

`\begin{gathered} A=3,000(1+(0.046)/(1))^1 \\ A=\$3,138 \end{gathered}`

therefore

Derek's annual interest is (3,138-3,000)=$138

Dereck's balance is $3,138

Is the same

For two years

t=2 years

simple interest

`\begin{gathered} A=3,000(1+0.046\cdot2) \\ A=\$3,276 \end{gathered}`

Suzanne's annual interest is (3,276-3,000)=$276

Suzanne's balance is $3,276

Compound interest

`\begin{gathered} A=3,000(1+(0.046)/(1))^2 \\ A=\$3,282.35 \end{gathered}`

interest is $282,35

Balance is $ 3,282.35